Symmetrical point with respect to a plane I have the point $A = (10,0,20)$, and I want the coordinates of its symmetrical point $B$ with respect to the plane $\pi = \begin{cases} x=2+3\alpha+\beta\\y=\alpha\\z=\alpha + 2\beta \end{cases}$, how can I do that? Look at the image:
image
These are the question's alternatives
$(2,20,24);\;(4,15,23);\;(-2,30,26);\;(0,25,25)$
 A: Long solution
The cartesian equation of the plane is $$\pi: 2x-5y-z-4=0$$
The parametric equations of the line $r$ which is perpendicular to $\pi$ and contains $A=(10,0,20)$ are $$r:\begin{cases}x=10+2t\\y=-5t\\z=20-t\end{cases}$$. This line also contains $B$ so the coordinates of $B$ are $(10+t,-5t,20-t)$ for some $t\in \Bbb R$.
The cartesian equations of $r$ are $$r:\begin{cases} y-5z+100=0 \\x+2z-50=0\end{cases}$$
The line $r$ intersects $\pi$ in the middle point $M$ between $A$ and $B$. Let's find the coordinates of $M$ by solving :$$\begin{cases} 2x-5y-z-4=0\\y-5z+100=0 \\x+2z-50=0\end{cases}$$
Once you've found the coordinates of $M=(x_M,y_M,z_M)$ you solve for $t$ $$||A-M||=||B-M||$$ $$\sqrt{(10-x_M)^2+(-y_M)^2+(20-z_M)^2}=\sqrt{(10+t-x_M)^2+(-5t-y_M)^2+(20-t-z_M)^2}$$
Lastly substitute the value of $t$ you've just found in $(10+t,-5t,20-t)$ to find the coordinates of $B$.
Shorter solution
For each alternative point $P$ that the exercise gives you, calculate the vector $\vec{AP}$ and check if it's parallel to the vector $(2,-5,-1)$ which is perpendicular to the plane. If this is true for more then one of the alternatives calculate the distance of $A$ and $P$ from $\pi$ and check if they're equal. The one alternative that verifies these two conditions ($\vec {AP}$ perpendicular to $\pi$ and $A$ and $P$ at the same distance from $\pi$) is $B$.
A: An equation of the plane is
\begin{equation}
x = 2 + 3 y + (z-y)/2 \Longleftrightarrow 2 x - 5 y -z -4 = 0
\end{equation}
Let $A = (x_0,y_0,z_0) = (10, 0, 20)$ and $B = (x_1, y_1, z_1)= A + t (2, -5, -1)$ because the vector $(2, -5, -1)$ is perpendicular to the plane. We have $2 x_0-5 y_0 -z_0 - 4 = -4$, hence we need $2 x_1-5 y_1 -z_1 - 4 = +4$ in order to have the midpoint $(A + B)/2$ on the plane. This implies $t = 8/30 = 4/ 15$, hence
\begin{equation}
x_1 = 10 + \frac{8}{15}=\frac{158}{15}\qquad
y_1 = -\frac{20}{15}\qquad z_1 = 20 -\frac{4}{15} = \frac{296}{15}
\end{equation}
